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Question 552062: Please help me solve,
What is an equation for the hyperbola with vertices (3,0) and (-3,0) and asymptote y=4/3x?
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! there's a lot to understand about hyperbolas that can be confusing.
not having seen them in a while, i forgot most of what i had previously learned.
a good reference to brush up on is:
http://www.purplemath.com/modules/hyperbola.htm
your problem states the following:
vertices are (3,0) and (-3,0) and asymptote is given by the equation y = (4/3)x.
this makes your hyperbola a horizontally aligned hyperbola.
the formula for a horizontally aligned hyperbola is
(x-h)^2/a^2 - (y-k)^2/b^2 = 1
the center of this hyperbola will be halfway between the left vertex and the right vertex.
that would be halfway between (-3,0) and (3,0) which is equal to (0,0).
this makes h = 0 and k = 0 and the formula becomes:
x^2/a^2 - y^2/b^2 = 1
the formula for the asymptote is y = (4/3)x
the general formula for the asymptote of a horizontally aligned hyperbola is:
y = (b/a)(x-h)+k
since h and k are both equal to 0, this becomes:
y = (b/a)x
we are given that y = (4/3)x.
this makes:
b = 4
a = 3
our equation becomes:
x^2/3^2 - y^2/4^2 = 1
this simplifies to:
x^2/9 - y^2/16 = 1
if we multiply both sides of this equation by 144, we get:
16x^2 - 9y^2 = 144
add 9y^2 to both sides of this equation and subtract 144 from both sides of this equation to get:
9y^2 = 16x^2 - 144
divide both sides of this equation by 9 to get:
y^2 = (16x^2-144)/9
take the square root of both sides of this equation to get:
y = +/- sqrt((16x^2-144)/9)
graph both these equations as shown below:
the equation for the asymptotes for this equation are:
y = +/- (4/3)x
add these to the graph to get:
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